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My question is:

Solve simultaneously:

$$\left\{\begin{align*} &\frac{xy}{x+y}=1\\ &\frac{xz}{x+z}=2\\ &\frac{yz}{y+z}=3 \end{align*}\right.$$

I am unable to solve this simultaneous equation. Any hints to solve this question would be of great help.

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up vote 1 down vote accepted

Clearly $x,y$, and $z$ are all non-zero: otherwise at least two of the fractions would be $0$. Thus, we may invert everything and rewrite the system as

$$\left\{\begin{align*} &\frac{x+y}{xy}=1\\\\ &\frac{x+z}{xz}=\frac12\\\\ &\frac{y+z}{yz}=\frac13\;. \end{align*}\right.$$

Now divide out on the lefthand side:

$$\left\{\begin{align*} &\frac1x+\frac1y=1\\\\ &\frac1x+\frac1z=\frac12\\\\ &\frac1y+\frac1z=\frac13\;. \end{align*}\right.$$

Now substitute $u=\dfrac1x,v=\dfrac1y$, and $w=\dfrac1z$ to get

$$\left\{\begin{align*} &u+v=1\\\\ &u+w=\frac12\\\\ &v+w=\frac13\;. \end{align*}\right.$$

This is a very easy system to solve, and once you’ve solved it, you can easily get $x,y$, and $z$.

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RHS should be $1, \tfrac{1}{2}, \tfrac{1}{3}$. – WimC Jun 2 '12 at 7:49
@WimC: Yes, I caught that as soon as I hit Post; I figured that that’s what the waiting comment would be. :-) Thanks. – Brian M. Scott Jun 2 '12 at 7:50

A more complicated approach:

As with Brian's solution, note that all $x,y,z \neq 0$.

Note that $z\neq2$, since if it was then the 2nd equation would give an immediate contradiction. Similarly, $z\neq 3$ using the 3rd equation.

Then the 2nd equation gives $x(z-2) = 2z$, or $x = \frac{2z}{z-2}$. The 3rd equation gives $y(z-3) = 3z$, or $y = \frac{3z}{z-3}$.

Substitution into the 1st equation gives: $$\frac{6z^2}{(z-2)(z-3)} = \frac{2z}{z-2} + \frac{3z}{z-3},$$ or equivalently, $z(z+12) = 0$. From this we get $z=-12$, $x = \frac{12}{7}$, $x = \frac{12}{5}$.

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